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2 years ago

How many moles of Gold (III) sulfide must decompose to produce 4.32 moles of gold

Chemistry

1 answer:

sertanlavr [38]2 years ago

3 0

Answer:

2.16 moles of Au₂S₃ are decomposed.

Explanation:

Given data:

Number of moles of gold (III) sulfide decomposed = ?

Number of moles of gold metal formed = 4.32 mol

Solution:

Chemical equation:

Au₂S₃ → 2Au + 3S

Now we will compare the moles of Au₂S₃ with gold metal.

Au : Au₂S₃

2 : 1

4.32 : 1/2×4.32 = 2.16

2.16 moles of Au₂S₃ are decomposed.

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4. A student purified a 500-mg sample of phthalic acid by recrystallization from water. The published solubility of phthalic aci

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Answer:

2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.

Explanation:

We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.

Now we devise the following reasoning:

If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C

Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C

X = (0.5 × 100) / 18 = 2.77 mL of water

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3 years ago

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I think it might just might be e

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3 years ago

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2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

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x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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2 years ago

Which statement describes the composition of potassium chlorate, KClO3?

Aliun [14]

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3 years ago

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